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7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 Work Principle of Virtual Work Deflections of Trusses by the Virtual Work Method Deflections of Beams by the Virtual Work Method Deflections of Frames by the Virtual Work Method Conservation of Energy and Strain Energy Castigliano’s Second Theorem Betti’s Law and Maxwell’s Law of Reciprocal Deflections Summary Problems Interstate 35W Bridge Collapse in Minnesota (2007) AP Photo/Pioneer Press, Brandi Jade Thomas In this chapter, we develop methods for the analysis of deﬂections of statically determinate structures by using some basic principles of work and energy. Work–energy methods are more general than the geometric methods considered in the previous chapter in the sense that they can be applied to various types of structures, such as trusses, beams, and frames. A disadvantage of these methods is that with each application, only one deﬂection component, or slope, at one point of the structure can be computed. We begin by reviewing the basic concept of work performed by forces and couples during a deformation of the structure and then discuss the principle of virtual work. This principle is used to formulate the method of virtual work for the deﬂections of trusses, beams, and frames. We derive the expressions for strain energy of trusses, beams, and frames and then consider Castigliano’s second theorem for computing deﬂections. Finally, we present Betti’s law and Maxwell’s law of reciprocal deﬂections. 275 276 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 7.1 WORK The work done by a force acting on a structure is simply deﬁned as the force times the displacement of its point of application in the direction of the force. Work is considered to be positive when the force and the displacement in the direction of the force have the same sense and negative when the force and the displacement have opposite sense. Let us consider the work done by a force P during the deformation of a structure under the action of a system of forces (which includes P), as shown in Fig. 7.1(a). The magnitude of P may vary as its point of application displaces from A in the undeformed position of the structure to A 0 in the ﬁnal deformed position. The work dW that P performs as its point of application undergoes an inﬁnitesimal displacement, dD (Fig. 7.1(a)), can be written as dW ¼ PðdDÞ The total work W that the force P performs over the entire displacement D is obtained by integrating the expression of dW as ðD P dD (7.1) W¼ 0 FIG. 7.1 SECTION 7.1 Work 277 As Eq. (7.1) indicates, the work is equal to the area under the force-displacement diagram as shown in Fig. 7.1(b). In this text, we are focusing our attention on the analysis of linear elastic structures, so an expression for work of special interest is for the case when the force varies linearly with displacement from zero to its ﬁnal value, as shown in Fig. 7.1(c). The work for such a case is given by the triangular area under the force-displacement diagram and is expressed as 1 W ¼ PD 2 (7.2) Another special case of interest is depicted in Fig. 7.1(d). In this case, the force remains constant at P while its point of application undergoes a displacement D caused by some other action independent of P. The work done by the force P in this case is equal to the rectangular area under the force-displacement diagram and is expressed as W ¼ PD (7.3) It is important to distinguish between the two expressions for work as given by Eqs. (7.2) and (7.3). Note that the expression for work for the case when the force varies linearly with displacement (Eq. 7.2) contains a factor of 12 , whereas the expression for work for the case of a constant force (Eq. 7.3) does not contain this factor. These two expressions for work will be used subsequently in developing di¤erent methods for computing deﬂections of structures. The expressions for the work of couples are similar in form to those for the work of forces. The work done by a couple acting on a structure is deﬁned as the moment of the couple times the angle through which the couple rotates. The work dW that a couple of moment M performs through an inﬁnitesimal rotation dy (see Fig. 7.1(a)) is given by dW ¼ MðdyÞ Therefore, the total work W of a couple with variable moment M over the entire rotation y can be expressed as ðy W ¼ M dy (7.4) 0 When the moment of the couple varies linearly with rotation from zero to its ﬁnal value, the work can be expressed as 1 W ¼ My 2 (7.5) and, if M remains constant during a rotation y, then the work is given by W ¼ My (7.6) 278 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 7.2 PRINCIPLE OF VIRTUAL WORK The principle of virtual work, which was introduced by John Bernoulli in 1717, provides a powerful analytical tool for many problems of structural mechanics. In this section, we study two formulations of this principle, namely, the principle of virtual displacements for rigid bodies and the principle of virtual forces for deformable bodies. The latter formulation is used in the following sections to develop the method of virtual work, which is considered to be one of the most general methods for determining deﬂections of structures. Principle of Virtual Displacements for Rigid Bodies The principle of virtual displacements for rigid bodies can be stated as follows: If a rigid body is in equilibrium under a system of forces and if it is subjected to any small virtual rigid-body displacement, the virtual work done by the external forces is zero. The term virtual simply means imaginary, not real. Consider the beam shown in Fig. 7.2(a). The free-body diagram of the beam is shown in Fig. 7.2(b), in which Px and Py represent the components of the external load P in the x and y directions, respectively. Now, suppose that the beam is given an arbitrary small virtual rigid-body displacement from its initial equilibrium position ABC to another position A 0 B 0 C 0 , as shown in Fig. 7.2(c). As shown in this ﬁgure, the total virtual rigid-body displacement of the beam can be decomposed into translations D vx and D vy in the x and y directions, respectively, and a rotation yv about point A. Note that the subscript v is used here to identify the displacements as virtual quantities. As the beam undergoes the virtual displacement from position ABC to position A 0 B 0 C 0 , the forces acting on it perform work, which is called virtual work. The total virtual work, Wve , performed by the external forces acting on the beam can be expressed as the sum of the virtual work Wvx and Wvy done during translations in the x and y directions, respectively, and the virtual work Wvr , done during the rotation; that is, Wve ¼ Wvx þ Wvy þ Wvr (7.7) During the virtual translations D vx and D vy of the beam, the virtual work done by the forces is given by P Wvx ¼ Ax D vx Px D vx ¼ ðAx Px ÞD vx ¼ ð Fx ÞD vx (7.8) and Wvy ¼ Ay D vy Py D vy þ Cy D vy ¼ ðAy Py þ Cy ÞD vy ¼ ð P Fy ÞD vy (7.9) SECTION 7.2 FIG. Principle of Virtual Work 279 7.2 (see Fig. 7.2(c)). The virtual work done by the forces during the small virtual rotation yv can be expressed as P Wvr ¼ Py ðayv Þ þ Cy ðLyv Þ ¼ ðaPy þ LCy Þyv ¼ ð MA Þyv (7.10) By substituting Eqs. (7.8) through (7.10) into Eq. (7.7), we write the total virtual work done as P P P Wve ¼ ð Fx ÞD vx þ ð Fy ÞD vy þ ð MA Þyv (7.11) P P Because the beam is in equilibrium, Fx ¼ 0, Fy ¼ 0, and P MA ¼ 0; therefore, Eq. (7.11) becomes Wve ¼ 0 (7.12) which is the mathematical statement of the principle of virtual displacements for rigid bodies. 280 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods Principle of Virtual Forces for Deformable Bodies The principle of virtual forces for deformable bodies can be stated as follows: If a deformable structure is in equilibrium under a virtual system of forces (and couples) and if it is subjected to any small real deformation consistent with the support and continuity conditions of the structure, then the virtual external work done by the virtual external forces (and couples) acting through the real external displacements (and rotations) is equal to the virtual internal work done by the virtual internal forces (and couples) acting through the real internal displacements (and rotations). In this statement, the term virtual is associated with the forces to indicate that the force system is arbitrary and does not depend on the action causing the real deformation. To demonstrate the validity of this principle, consider the twomember truss shown in Fig. 7.3(a). The truss is in equilibrium under the action of a virtual external force Pv as shown. The free-body diagram of joint C of the truss is shown in Fig. 7.3(b). Since joint C is in equilibrium, the virtual external and internal forces acting on it must satisfy the following two equilibrium equations: P Fx ¼ 0 Pv FvAC cos y1 FvBC cos y2 ¼ 0 (7.13) P FvAC sin y1 þ FvBC sin y2 ¼ 0 Fy ¼ 0 in which FvAC and FvBC represent the virtual internal forces in members AC and BC, respectively, and y1 and y2 denote, respectively, the angles of inclination of these members with respect to the horizontal (Fig. 7.3(a)). Now, let us assume that joint C of the truss is given a small real displacement, D, to the right from its equilibrium position, as shown in Fig. 7.3(a). Note that the deformation is consistent with the support FIG. 7.3 SECTION 7.2 Principle of Virtual Work 281 conditions of the truss; that is, joints A and B, which are attached to supports, are not displaced. Because the virtual forces acting at joints A and B do not perform any work, the total virtual work for the truss ðWv Þ is equal to the algebraic sum of the work of the virtual forces acting at joint C; that is, Wv ¼ Pv D FvAC ðD cos y1 Þ FvBC ðD cos y2 Þ or Wv ¼ ðPv FvAC cos y1 FvBC cos y2 ÞD (7.14) As indicated by Eq. (7.13), the term in the parentheses on the right-hand side of Eq. (7.14) is zero; therefore, the total virtual work is Wv ¼ 0. Thus, Eq. (7.14) can be expressed as Pv D ¼ FvAC ðD cos y1 Þ þ FvBC ðD cos y2 Þ (7.15) in which the quantity on the left-hand side represents the virtual external work ðWve Þ done by the virtual external force, Pv , acting through the real external displacement, D. Also, realizing that the terms D cos y1 and D cos y2 are equal to the real internal displacements (elongations) of members AC and BC, respectively, we can conclude that the right-hand side of Eq. (7.15) represents the virtual internal work ðWvi Þ done by the virtual internal forces acting through the real internal displacements; that is Wve ¼ Wvi (7.16) which is the mathematical statement of the principle of virtual forces for deformable bodies. It should be realized that the principle of virtual forces as described here is applicable regardless of the cause of real deformations; that is, deformations due to loads, temperature changes, or any other e¤ect can be determined by the application of the principle. However, the deformations must be small enough so that the virtual forces remain constant in magnitude and direction while performing the virtual work. Also, although the application of this principle in this text is limited to elastic structures, the principle is valid regardless of whether the structure is elastic or not. The method of virtual work is based on the principle of virtual forces for deformable bodies as expressed by Eq. (7.16), which can be rewritten as virtual external work ¼ virtual internal work or, more speciﬁcally, as (7.17) 282 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods Virtual system P virtual external force P virtual internal force ¼ real external displacement real internal displacement Real system (7.18) in which the terms forces and displacements are used in a general sense and include moments and rotations, respectively. Note that because the virtual forces are independent of the actions causing the real deformation and remain constant during the real deformation, the expressions of the external and internal virtual work in Eq. (7.18) do not contain the factor 1/2. As Eq. (7.18) indicates, the method of virtual work employs two separate systems: a virtual force system and the real system of loads (or other e¤ects) that cause the deformation to be determined. To determine the deﬂection (or slope) at any point of a structure, a virtual force system is selected so that the desired deﬂection (or rotation) will be the only unknown in Eq. (7.18). The explicit expressions of the virtual work method to be used for computing deﬂections of trusses, beams, and frames are developed in the following three sections. 7.3 DEFLECTIONS OF TRUSSES BY THE VIRTUAL WORK METHOD To develop the expression of the virtual work method that can be used to determine the deﬂections of trusses, consider an arbitrary statically determinate truss, as shown in Fig. 7.4(a). Let us assume that we want to determine the vertical deﬂection, D, at joint B of the truss due to the given external loads P1 and P2 . The truss is statically determinate, so the axial forces in its members can be determined from the method of joints described previously in Chapter 4. If F represents the axial force in an arbitrary member j (e.g., member CD in Fig. 7.4(a)) of the truss, then (from mechanics of materials) the axial deformation, d, of this member is given by d¼ FL AE (7.19) in which L; A, and E denote, respectively, the length, cross-sectional area, and modulus of elasticity of member j. To determine the vertical deﬂection, D, at joint B of the truss, we select a virtual system consisting of a unit load acting at the joint and in the direction of the desired deﬂection, as shown in Fig. 7.4(b). Note SECTION 7.3 Deflections of Trusses by the Virtual Work Method 283 that the (downward) sense of the unit load in Fig. 7.4(b) is the same as the assumed sense of the desired deﬂection D in Fig. 7.4(a). The forces in the truss members due to the virtual unit load can be determined from the method of joints. Let Fv denote the virtual force in member j. Next, we subject the truss with the virtual unit load acting on it (Fig. 7.4(b)) to the deformations of the real loads (Fig. 7.4(a)). The virtual external work performed by the virtual unit load as it goes through the real deﬂection D is equal to Wve ¼ 1ðDÞ (7.20) To determine the virtual internal work, let us focus our attention on member j (member CD in Fig. 7.4). The virtual internal work done on member j by the virtual axial force Fv , acting through the real axial deformation d, is equal to Fv d. Therefore, the total virtual internal work done on all the members of the truss can be written as P (7.21) Wvi ¼ Fv ðdÞ By equating the virtual external work (Eq. (7.20)) to the virtual internal work (Eq. (7.21)) in accordance with the principle of virtual forces for deformable bodies, we obtain the following expression for the method of virtual work for truss deﬂections: FIG. 7.4 1ðDÞ ¼ P Fv ðdÞ (7.22) When the deformations are caused by external loads, Eq. (7.19) can be substituted into Eq. (7.22) to obtain 1ðDÞ ¼ P Fv FL AE (7.23) Because the desired deﬂection, D, is the only unknown in Eq. (7.23), its value can be determined by solving this equation. Temperature Changes and Fabrication Errors The expression of the virtual work method as given by Eq. (7.22) is quite general in the sense that it can be used to determine truss deﬂections due to temperature changes, fabrication errors, and any other e¤ect for which the member axial deformations, d, are either known or can be evaluated beforehand. The axial deformation of a truss member j of length L due to a change in temperature ðDTÞ is given by d ¼ aðDTÞL (7.24) 284 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods in which a denotes the coe‰cient of thermal expansion of member j. Substituting Eq. (7.24) into Eq. (7.22), we obtain the following expression: 1ðDÞ ¼ P Fv aðDTÞL (7.25) which can be used to compute truss deﬂections due to the changes in temperature. Truss deﬂections due to fabrication errors can be determined by simply substituting changes in member lengths due to fabrication errors for d in Eq. (7.22). Procedure for Analysis The following step-by-step procedure can be used to determine the deﬂections of trusses by the virtual work method. 1. Real System If the deﬂection of the truss to be determined is caused by external loads, then apply the method of joints and/or the method of sections to compute the (real) axial forces ðF Þ in all the members of the truss. In the examples given at the end of this section, tensile member forces are considered to be positive and vice versa. Similarly, increases in temperature and increases in member lengths due to fabrication errors are considered to be positive and vice versa. 2. Virtual System Remove all the given (real) loads from the truss; then apply a unit load at the joint where the deﬂection is desired and in the direction of the desired deﬂection to form the virtual force system. By using the method of joints and/or the method of sections, compute the virtual axial forces ðFv Þ in all the members of the truss. The sign convention used for the virtual forces must be the same as that adopted for the real forces in step 1; that is, if real tensile forces, temperature increases, or member elongations due to fabrication errors were considered as positive in step 1, then the virtual tensile forces must also be considered to be positive and vice versa. 3. The desired deﬂection of the truss can now be determined by applying Eq. (7.23) if the deﬂection is due to external loads, Eq. (7.25) if the deﬂection is caused by temperature changes, or Eq. (7.22) in the case of the deﬂection due to fabrication errors. The application of these virtual work expressions can be facilitated by arranging the real and virtual quantities, computed in steps 1 and 2, in a tabular form, as illustrated in the following examples. A positive answer for the desired deﬂection means that the deﬂection occurs in the same direction as the unit load, whereas a negative answer indicates that the deﬂection occurs in the direction opposite to that of the unit load. SECTION 7.3 Deflections of Trusses by the Virtual Work Method 285 Example 7.1 Determine the horizontal deﬂection at joint C of the truss shown in Fig. 7.5(a) by the virtual work method. Solution Real System The real system consists of the loading given in the problem, as shown in Fig. 7.5(b). The member axial forces due to the real loads ðF Þ obtained by using the method of joints are also depicted in Fig. 7.5(b). Virtual System The virtual system consists of a unit (1 kN) load applied in the horizontal direction at joint C, as shown in Fig. 7.5(c). The member axial forces due to the 1 kN virtual load ðFv Þ are determined by applying the method of joints. These member forces are also shown in Fig. 7.5(c). Horizontal Deﬂection at C, D C To facilitate the computation of the desired deﬂection, the real and virtual member forces are tabulated along with the member lengths ðLÞ, as shown in Table 7.1. As the values of the cross-sectional area, A, and modulus of elasticity, E, are the same for all the members, these are not included in the table. Note that the same sign convention is used for both real and virtual systems; that is, in both the third and the fourth columns of the table, tensile forces are entered as positive numbers and compressive forces as negative numbers. Then, for each member, the quantity Fv ðFLÞ is computed, and its value is entered in the ﬁfth column of the table. P The algebraic sum of all of the entries in the ﬁfth column, Fv ðFLÞ, is then determined, and its value is recorded at the bottom of the ﬁfth column, as shown. The total virtual internal work done on all of the members of the truss is given by Wvi ¼ 1 P Fv ðFLÞ EA C C 200 kN 200 kN 487 .5 31 2.5 3.6 m A FIG. 7.5 B 1.2m 1.5m EA = constant E = 70 Gpa A = 40 cm2 (a) A 250 187.5 B 450 (b) Real System — F Forces continued 286 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 1 KN 3.2 5 3.7 5 C 1 A 3 FIG. 1.25 B 3 (c) Virtual System — Fv Forces 7.5 (contd.) The virtual external work done by the 1 kN load acting through the desired horizontal deﬂection at C, D C , is Wve ¼ ð1kNÞD C Finally, we determine the desired deﬂection D C by equating the virtual external work to the virtual internal work and solving the resulting equation for D C as shown in Table 7.1. Note that the positive answer for D C indicates that joint C deﬂects to the right, in the direction of the unit load. TABLE 7.1 Member AB AC BC L (m) F (kN) Fv (kN) Fv ðFLÞ (kN 2 m) 1.2 4.5 3.9 187.5 312.5 487.5 1.25 3.75 3.25 281.25 5273.44 6179.06 P 1ðD C Þ ¼ ð1 kNÞD C ¼ Fv ðFLÞ ¼ 11733:75 1 P Fv ðFLÞ EA 11733:75 kN m 70ð10 6 Þ 4000ð10 6 Þ D C ¼ 0:042 m D C ¼ 42 mm ! Ans. SECTION 7.3 Deflections of Trusses by the Virtual Work Method 287 Example 7.2 Determine the horizontal deﬂection at joint G of the truss shown in Fig. 7.6(a) by the virtual work method. FIG. 7.6 Solution Real System The real system consists of the loading given in the problem, as shown in Fig. 7.6(b). The member axial forces due to the real loads ðF Þ obtained by using the method of joints are also shown in Fig. 7.6(b). Virtual System The virtual system consists of a unit (1-kN) load applied in the horizontal direction at joint G, as shown in Fig. 7.6(c). The member axial forces due to the 1 kN virtual load ðFv Þ are also depicted in Fig. 7.6(c). Horizontal Deﬂection at G, DG To facilitate the computation of the desired deﬂection, the real and virtual member forces are tabulated along with the lengths ðLÞ and the cross-sectional areas ðAÞ of the members, as shown in Table 7.2. The modulus of elasticity, E, is the same for all the members, so its value is not included in the table. Note that the same sign convention is used for both real and virtual systems; that is, in both the fourth and the ﬁfth columns of the table, tensile forces are entered as positive numbers, and compressive forces as negative numbers. Then, for each member the quantity Fv ðFL=AÞ is computed, and its value is entered in the sixth column of the table. The algebraic sum of all the P entries in the sixth column, Fv ðFL=AÞ, is then determined, and its value is recorded at the bottom of the sixth column, as shown. Finally, the desired deﬂection DG is determined by applying the virtual work expression (Eq. (7.23)) as shown in Table 7.2. Note that the positive answer for DG indicates that joint G deﬂects to the right, in the direction of the unit load. continued 288 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods TABLE 7.2 Member AB CD EG AC CE BD DG BC CG L (m) A (m 2 ) F (kN) Fv (kN) Fv ðFL=AÞ (kN 2 /m) 4 4 4 3 3 3 3 5 5 0.003 0.002 0.002 0.003 0.003 0.003 0.003 0.002 0.002 300 0 100 300 0 75 75 375 125 1 0 0 1.5 0 0.75 0.75 1.25 1.25 400000 0 0 450000 0 56250 56250 1171875 390625 P Fv FL ¼ 2525000 A 1P FL Fv 1ðD G Þ ¼ E A ð1 kNÞD G ¼ 2525000 200ð106 Þ D G ¼ 0:0126 m D G ¼ 12:6 mm ! Ans. Example 7.3 Determine the horizontal and vertical components of the deﬂection at joint B of the truss shown in Fig. 7.7(a) by the virtual work method. FIG. 7.7 continued SECTION 7.3 FIG. Deflections of Trusses by the Virtual Work Method 289 7.7 (contd.) Solution Real System The real system and the corresponding member axial forces ðF Þ are shown in Fig. 7.7(b). Horizontal Deﬂection at B, DBH The virtual system used for determining the horizontal deﬂection at B consists of a 1-kN load applied in the horizontal direction at joint B, as shown in Fig. 7.7(c). The member axial forces ðFv1 Þ due to this virtual load are also shown in this ﬁgure. The member axial forces due to the real system ðF Þ and this virtual system ðFv1 Þ are then tabulated, and the virtual work expression given by Eq. (7.23) is applied to determine DBH , as shown in Table 7.3. TABLE 7.3 Member AB BC AD BD CD 1ðDBH Þ ¼ ð1 kNÞDBH ¼ L (m) F (kN) Fv1 (kN) Fv1 ðFLÞ (kN 2m) Fv2 (kN) Fv2 ðFLÞ (kN 2m) 4 3 5.66 4 5 P 21 21 79.2 84 35 1 0 0 0 0 84 0 0 0 0 0.43 0.43 0.61 1 0.71 36.12 27.09 273.45 336.00 124.25 Fv ðFLÞ 84 1 P Fv1 ðFLÞ EA 1ðDBV Þ ¼ 84 kN m 200ð10 6 Þð0:0012Þ kNm DBH ¼ 0:00035 m DBH ¼ 0:35 mm ! ð1 kNÞDBV ¼ 796.91 1 P Fv2 ðFLÞ EA 796:91 kN m 200ð10 6 Þð0:0012Þ kNm DBV ¼ 0:00332 m Ans. DBV ¼ 3:32 mm # Ans. continued 290 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods Vertical Deﬂection at B, DBV The virtual system used for determining the vertical deﬂection at B consists of a 1-kN load applied in the vertical direction at joint B, as shown in Fig. 7.7(d). The member axial forces ðFv2 Þ due to this virtual load are also shown in this ﬁgure. These member forces are tabulated in the sixth column of Table 7.3, and DBV is computed by applying the virtual work expression (Eq. (7.23)), as shown in the table. Example 7.4 Determine the vertical deﬂection at joint C of the truss shown in Fig. 7.8(a) due to a temperature drop of 8 C in members AB and BC and a temperature increase of 30 C in members AF ; FG; GH, and EH. Use the virtual work method. Solution Real System The real system consists of the temperature changes ðDTÞ given in the problem, as shown in Fig. 7.8(b). Virtual System The virtual system consists of a 1-kN load applied in the vertical direction at joint C, as shown in Fig. 7.8(c). Note that the virtual axial forces ðFv Þ are computed for only those members that are subjected to temperature changes. Because the temperature changes in the remaining members of the truss are zero, their axial deformations are zero; therefore, no internal virtual work is done on those members. FIG. 7.8 continued SECTION 7.3 Deflections of Trusses by the Virtual Work Method 291 Vertical Deﬂection at C, D C The temperature changes ðDTÞ and the virtual member forces ðFv Þ are tabulated along with the lengths ðLÞ of the members, in Table 7.4. The coe‰cient of thermal expansion, a, is the same for all the members, so its value is not included in the table. The desired deﬂection D C is determined by applying the virtual work expression given by Eq. (7.25), as shown in the table. Note that the negative answer for D C indicates that joint C deﬂects upward, in the direction opposite to that of the unit load. TABLE 7.4 Member AB BC AF FG GH EH L (m) DT ( C) Fv (kN) Fv ðDTÞL (kN- C-m) 3 3 3.75 3.75 3.75 3.75 8 8 30 30 30 30 0.667 0.667 0.833 0.833 0.833 0.833 16.0 16.0 93.7 93.7 93.7 93.7 P Fv ðDTÞL ¼ 406:8 P 1ðD C Þ ¼ a Fv ðDTÞL ð1 kNÞD C ¼ 1:2ð105 Þð406:8Þ D C ¼ 0:00488 m D C ¼ 4:88 mm " Ans. Example 7.5 Determine the vertical deﬂection at joint D of the truss shown in Fig. 7.9(a) if member CF is 15 mm too long and member EF is 10 mm too short. Use the method of virtual work. Solution Real System The real system consists of the changes in the lengths ðdÞ of members CF and EF of the truss, as shown in Fig. 7.9(b). Virtual System The virtual system consists of a 1-kN load applied in the vertical direction at joint D, as shown in Fig. 7.9(c). The necessary virtual forces ðFv Þ in members CF and EF can be easily computed by using the method of sections. Vertical Deﬂection at D, DD The desired deﬂection is determined by applying the virtual work expression given by Eq. (7.22), as shown in Table 7.5. continued 292 FIG. CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 7.9 TABLE 7.5 Member CF EF d (mm) Fv (kN) Fv ðdÞ (kN-mm) 15 10 1 1 15 10 P 1ðDD Þ ¼ P Fv ðdÞ ¼ 25 Fv ðdÞ ð1 kNÞDD ¼ 25 kN-mm DD ¼ 25 mm DD ¼ 25 mm " Ans. 312 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods Therefore, DC ¼ ¼ 418:125 kN2 -m 3246:75 kN2 -m þ AE EI 418:125 3246:75 þ 225ð104 Þ200ð106 Þ 200ð106 Þ400ð106 Þ ¼ 0:000093 þ 0:040584 ¼ 0:04068 m Ans. D C ¼ 40:68 mm: ! Note that the magnitude of the axial deformation term is negligibly small as compared to that of the bending deformation term. 7.6 CONSERVATION OF ENERGY AND STRAIN ENERGY Before we can develop the next method for computing deﬂections of structures, it is necessary to understand the concepts of conservation of energy and strain energy. The energy of a structure can be simply deﬁned as its capacity for doing work. The term strain energy is attributed to the energy that a structure has because of its deformation. The relationship between the work and strain energy of a structure is based on the principle of conservation of energy, which can be stated as follows: The work performed on an elastic structure in equilibrium by statically (gradually) applied external forces is equal to the work done by internal forces, or the strain energy stored in the structure. This principle can be mathematically expressed as W e ¼ Wi (7.39) We ¼ U (7.40) or In these equations, We and Wi represent the work done by the external and internal forces, respectively, and U denotes the strain energy of the structure. The explicit expression for the strain energy of a structure depends on the types of internal forces that can develop in the members of the structure. Such expressions for the strain energy of trusses, beams, and frames are derived in the following. SECTION 7.6 Conservation of Energy and Strain Energy 313 Strain Energy of Trusses Consider the arbitrary truss shown in Fig. 7.18. The truss is subjected to a load P, which increases gradually from zero to its ﬁnal value, causing the structure to deform as shown in the ﬁgure. Because we are considering linearly elastic structures, the deﬂection of the truss D at the point of application of P increases linearly with the load; therefore, as discussed in Section 7.1 (see Fig. 7.1(c)), the external work performed by P during the deformation D can be expressed as FIG. 7.18 1 We ¼ PD 2 To develop the expression for internal work or strain energy of the truss, let us focus our attention on an arbitrary member j (e.g., member CD in Fig. 7.18) of the truss. If F represents the axial force in this member due to the external load P, then as discussed in Section 7.3, the axial deformation of this member is given by d ¼ ðFLÞ=ðAEÞ. Therefore, internal work or strain energy stored in member j, Uj , is given by 1 F 2L Uj ¼ F d ¼ 2 2AE The strain energy of the entire truss is simply equal to the sum of the strain energies of all of its members and can be expressed as U¼ P F 2L 2AE (7.41) Note that a factor of 12 appears in the expression for strain energy because the axial force F and the axial deformation d caused by F in each member of the truss are related by the linear relationship d ¼ ðFLÞ=ðAEÞ. Strain Energy of Beams To develop the expression for the strain energy of beams, consider an arbitrary beam, as shown in Fig. 7.19(a). As the external load P acting on the beam increases gradually from zero to its ﬁnal value, the internal bending moment M acting on a di¤erential element dx of the beam (Fig. 7.19(a) and (b)) also increases gradually from zero to its ﬁnal value, while the cross sections of element dx rotate by an angle dy with respect to each other. The internal work or the strain energy stored in the element dx is, therefore, given by 1 dU ¼ MðdyÞ 2 (7.42) 314 FIG. CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 7.19 Recalling from Section 7.4 (Eq. (7.27)) that the change in slope, dy, can be expressed in terms of the bending moment, M, by the relationship dy ¼ ðM=EI Þ dx, we write Eq. (7.42) as dU ¼ M2 dx 2EI (7.43) The expression for the strain energy of the entire beam can now be obtained by integrating Eq. (7.43) over the length L of the beam: U¼ ðL 0 M2 dx 2EI (7.44) When the quantity M=EI is not a continuous function of x over the entire length of the beam, then the beam must be divided into segments so that M=EI is continuous in each segment. The integral on the righthand side of Eq. (7.44) is then evaluated by summing the integrals for all the segments of the beam. We must realize that Eq. (7.44) is based on the consideration of bending deformations of beams and does not include the e¤ect of shear deformations, which, as stated previously, are negligibly small as compared to the bending deformations for most beams. Strain Energy of Frames The portions of frames may be subjected to axial forces as well as bending moments, so the total strain energy (U) of frames is expressed SECTION 7.6 Conservation of Energy and Strain Energy 315 as the sum of the strain energy due to axial forces (Ua ) and the strain energy due to bending (Ub ); that is, U ¼ Ua þ Ub (7.45) If a frame is divided into segments so that the quantity F =AE is constant over the length L of each segment, then—as shown previously in the case of trusses—the strain energy stored in each segment due to the axial force F is equal to ðF 2 LÞ=ð2AEÞ. Therefore, the strain energy due to axial forces for the entire frame can be expressed as Ua ¼ P F 2L 2AE (7.46) Similarly, if the frame is divided into segments so that the quantity M=EI is continuous over each segment, then the strain energy stored in each segment due to bending can be obtained by integrating the quantity M=EI over the length of the segment (Eq. (7.44)). The strain energy due to bending for the entire frame is equal to the sum of strain energies of bending of all the segments of the frame and can be expressed as Ub ¼ ð P M2 dx 2EI (7.47) By substituting Eqs. (7.46) and (7.47) into Eq. (7.45), we obtain the following expression for the strain energy of frames due to both the axial forces and bending: U¼ ð P F 2L P M 2 þ dx 2AE 2EI (7.48) As stated previously, the axial deformations of frames are generally much smaller than the bending deformations and are usually neglected in the analysis. The strain energy of frames due only to bending is expressed as U¼ ð P M2 dx 2EI (7.49) 316 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 7.7 CASTIGLIANO’S SECOND THEOREM In this section, we consider another energy method for determining deﬂections of structures. This method, which can be applied only to linearly elastic structures, was initially presented by Alberto Castigliano in 1873 and is commonly known as Castigliano’s second theorem. (Castigliano’s ﬁrst theorem, which can be used to establish equations of equilibrium of structures, is not considered in this text.) Castigliano’s second theorem can be stated as follows: For linearly elastic structures, the partial derivative of the strain energy with respect to an applied force (or couple) is equal to the displacement (or rotation) of the force (or couple) along its line of action. In mathematical form, this theorem can be stated as: qU ¼ Di qPi or qU ¼ yi qM i (7.50) in which U ¼ strain energy; Di ¼ deﬂection of the point of application of the force Pi in the direction of Pi ; and yi ¼ rotation of the point of application of the couple M i in the direction of M i . To prove this theorem, consider the beam shown in Fig. 7.20. The beam is subjected to external loads P1 ; P2 , and P3 , which increase gradually from zero to their ﬁnal values, causing the beam to deﬂect, as shown in the ﬁgure. The strain energy (U) stored in the beam due to the external work (We ) performed by these forces is given by 1 1 1 U ¼ We ¼ P1 D1 þ P2 D2 þ P3 D3 2 2 2 (7.51) in which D1 ; D2 , and D3 are the deﬂections of the beam at the points of application of P1 ; P2 , and P3 , respectively, as shown in the ﬁgure. As Eq. (7.51) indicates, the strain energy U is a function of the external loads and can be expressed as U ¼ f ðP1 ; P2 ; P3 Þ (7.52) Now, assume that the deﬂection D2 of the beam at the point of application of P2 is to be determined. If P2 is increased by an inﬁnitesimal FIG. 7.20 SECTION 7.7 Castigliano’s Second Theorem 317 amount dP2 , then the increase in strain energy of the beam due to the application of dP2 can be written as dU ¼ qU dP2 qP2 (7.53) and the total strain energy, UT , now stored in the beam is given by UT ¼ U þ dU ¼ U þ qU dP2 qP2 (7.54) The beam is assumed to be composed of linearly elastic material, so regardless of the sequence in which the loads P1 ; ðP2 þ dP2 Þ, and P3 are applied, the total strain energy stored in the beam should be the same. Consider, for example, the sequence in which dP2 is applied to the beam before the application of P1 ; P2 , and P3 . If dD2 is the deﬂection of the beam at the point of application of dP2 due to dP2 , then the strain energy stored in the beam is given by ð1=2ÞðdP2 ÞðdD2 Þ. The loads P1 ; P2 , and P3 are then applied to the beam, causing the additional deﬂections D1 ; D2 , and D3 , respectively, at their points of application. Note that since the beam is linearly elastic, the loads P1 ; P2 , and P3 cause the same deﬂections, D1 ; D2 , and D3 , respectively, and perform the same amount of external work on the beam regardless of whether any other load is acting on the beam or not. The total strain energy stored in the beam during the application of dP2 followed by P1 ; P2 , and P3 is given by 1 1 1 1 UT ¼ ðdP2 ÞðdD2 Þ þ dP2 ðD2 Þ þ P1 D1 þ P2 D2 þ P3 D3 2 2 2 2 (7.55) Since dP2 remains constant during the additional deﬂection, D2 , of its point of application, the term dP2 ðD2 Þ on the right-hand side of Eq. (7.55) does not contain the factor 1=2. The term ð1=2ÞðdP2 ÞðdD2 Þ represents a small quantity of second order, so it can be neglected, and Eq. (7.55) can be written as 1 1 1 UT ¼ dP2 ðD2 Þ þ P1 D1 þ P2 D2 þ P3 D3 2 2 2 (7.56) By substituting Eq. (7.51) into Eq. (7.56) we obtain UT ¼ dP2 ðD2 Þ þ U (7.57) and by equating Eqs. (7.54) and (7.57), we write Uþ qU dP2 ¼ dP2 ðD2 Þ þ U qP2 or qU ¼ D2 qP2 which is the mathematical statement of Castigliano’s second theorem. 318 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods Application to Trusses To develop the expression of Castigliano’s second theorem, which can be used to determine the deﬂections of trusses, we substitute Eq. (7.41) for the strain energy (U) of trusses into the general expression of Castigliano’s second theorem for deﬂections as given by Eq. (7.50) to obtain D¼ q P F 2L qP 2AE (7.58) As the partial derivative qF 2 =qP ¼ 2F ðqF =qPÞ, the expression of Castigliano’s second theorem for trusses can be written as P qF FL D¼ qP AE (7.59) The foregoing expression is similar in form to the expression of the method of virtual work for trusses (Eq. (7.23)). As illustrated by the solved examples at the end of this section, the procedure for computing deﬂections by Castigliano’s second theorem is also similar to that of the virtual work method. Application to Beams By substituting Eq. (7.44) for the strain energy (U) of beams into the general expressions of Castigliano’s second theorem (Eq. (7.50)), we obtain the following expressions for the deﬂections and rotations, respectively, of beams: q D¼ qP ðL 0 M2 dx 2EI and q y¼ qM ðL 0 M2 dx 2EI or ð L D¼ 0 qM M dx qP EI (7.60) and y¼ ð L 0 qM M dx qM EI (7.61) SECTION 7.7 Castigliano’s Second Theorem 319 Application to Frames Similarly, by substituting Eq. (7.48) for the strain energy (U) of frames due to the axial forces and bending into the general expressions of Castigliano’s second theorem (Eq. (7.50)), we obtain the following expressions for the deﬂections and rotations, respectively, of frames: ð P qF FL P qM M þ dx D¼ qP AE qP EI (7.62) and ð P qF FL P qM M þ dx y¼ qM AE qM EI (7.63) When the e¤ect of axial deformations of the members of frames is neglected in the analysis, Eqs. (7.62) and (7.63) reduce to D¼ P ð qM M dx qP EI P ð qM M dx qM EI (7.64) and y¼ (7.65) Procedure for Analysis As stated previously, the procedure for computing deﬂections of structures by Castigliano’s second theorem is similar to that of the virtual work method. The procedure essentially involves the following steps. 1. 2. 3. If an external load (or couple) is acting on the given structure at the point and in the direction of the desired deﬂection (or rotation), then designate that load (or couple) as the variable P (or M ) and go to step 2. Otherwise, apply a ﬁctitious load P (or couple M ) at the point and in the direction of the desired deﬂection (or rotation). Determine the axial force F and/or the equation(s) for bending moment MðxÞ in each member of the structure in terms of P (or M ). Di¤erentiate the member axial forces F and/or the bending moments MðxÞ obtained in step 2 with respect to the variable P (or M ) to compute qF =qP and/or qM=qP (or qF =qM and/or qM=qM ). 320 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods 4. 5. Substitute the numerical value of P (or M ) into the expressions of F and/or MðxÞ and their partial derivatives. If P (or M ) represents a ﬁctitious load (or couple), its numerical value is zero. Apply the appropriate expression of Castigliano’s second theorem (Eqs. (7.59) through (7.65)) to determine the desired deﬂection or rotation of the structure. A positive answer for the desired deﬂection (or rotation) indicates that the deﬂection (or rotation) occurs in the same direction as P (or M ) and vice versa. Example 7.13 Determine the deﬂection at point C of the beam shown in Fig. 7.21(a) by Castigliano’s second theorem. FIG. 7.21 Solution This beam was previously analyzed by the moment-area, the conjugate-beam, and the virtual work methods in Examples 6.7, 6.13, and 7.9, respectively. The 60-kN external load is already acting at point C, where the deﬂection is to be determined, so we designate this load as the variable P, as shown in Fig. 7.21(b). Next, we compute the reactions of the beam in terms of P. These are also shown in Fig. 7.21(b). Since the loading is discontinuous at point B, the beam is divided into two segments, AB and BC. The x coordinates used for determining the equations for the bending moment in the two segments of the beam are shown in Fig. 7.21(b). The equations for M (in terms of P) obtained for the segments of the beam are tabulated in Table 7.11, along with the partial derivatives of M with respect to P. continued SECTION 7.7 Castigliano’s Second Theorem 321 TABLE 7.11 x Coordinate Segment Origin Limits (m) AB A 0–9 CB C 0–3 qM (kN-m/kN) qP M (kN-m) P 135 x x2 3 Px x 3 x The deﬂection at C can now be determined by substituting P ¼ 60 kN into the equations for M and qM=qP and by applying the expression of Castigliano’s second theorem as given by Eq. (7.60): ð L qM M DC ¼ dx qP EI 0 ð 9 ð3 1 x 60x 135x x 2 dx þ ðxÞð60xÞ dx DC ¼ EI 0 3 3 0 ð 9 ð3 1 x 2 ¼ ð115x x Þ dx þ ðxÞð60xÞ dx EI 0 3 0 ¼ 8228:25 kN-m 3 8228:25 ¼ ¼ 0:0514 m EI 200ð106 Þ800ð106 Þ The negative answer for D C indicates that point C deﬂects upward in the direction opposite to that of P. D C ¼ 51:4 mm " Ans. Example 7.14 Use Castigliano’s second theorem to determine the deﬂection at point B of the beam shown in Fig. 7.22(a). P B A L EI = constant (a) P B A x FIG. 7.22 (b) continued 322 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods Solution Using the x coordinate shown in Fig. 7.22(b), we write the equation for the bending moment in the beam as M ¼ Px The partial derivative of M with respect to P is given by qM ¼ x qP The deﬂection at B can now be obtained by applying the expression of Castigliano’s second theorem, as given by Eq. (7.60), as follows: ð L qM M dx DB ¼ qP EI 0 ðL Px DB ¼ ðxÞ dx EI 0 ð P L 2 PL 3 x dx ¼ ¼ 3EI EI 0 DB ¼ PL 3 # 3EI Ans. Example 7.15 Determine the rotation of joint C of the frame shown in Fig. 7.23(a) by Castigliano’s second theorem. Solution This frame was previously analyzed by the virtual work method in Example 7.10. No external couple is acting at joint C, where the rotation is desired, so we apply a ﬁctitious couple M ð¼ 0Þ at C, as shown in Fig. 7.23(b). The x coordinates used for determining the bending moment equations for the three segments of the frame are also shown in Fig. 7.23(b), and the equations for M in terms of M and qM=qM obtained for the three segments are tabulated in Table 7.12. The rotation of joint C of the frame can now be determined by setting M ¼ 0 in the equations for M and qM=qM and by applying the expression of Castigliano’s second theorem as given by Eq. (7.65): ð P qM M dx yC ¼ qM EI ð 12 x x2 dx ¼ 180x 20 2 12 0 ¼ 4320 kN-m 2 4320 ¼ ¼ 0:0216 rad EI 200ð106 Þ1000ð106 Þ yC ¼ 0:0216 rad @ Ans. continued SECTION 7.7 FIG. Castigliano’s Second Theorem 7.23 TABLE 7.12 Origin Limits (m) M (kN-m) qM kN-m qM kN-m AB CB A C 0–4 0–4 DC D 0–12 180x 720 M x2 x 20 180 þ 12 2 0 0 x 12 x Coordinate Segment 323 324 CHAPTER 7 Deflections of Trusses, Beams, and Frames: Work–Energy Methods Example 7.16 Use Castigliano’s second theorem to determine the horizontal and vertical components of the deﬂection at joint B of the truss shown in Fig. 7.24(a). FIG. 7.24 Solution This truss was previously analyzed by the virtual work method in Example 7.3. TABLE 7.13 Member L (m) F (kN) qF qP1 (kN/kN) AB BC AD BD CD 4 3 5.66 4 5 15 þ P1 þ 0:43P2 15 þ 0:43P2 28:28 0:61P2 P2 25 0:71P2 1 0 0 0 0 For P1 ¼ 0 and P2 ¼ 84 kN qF qP2 (kN/kN) ðqF =qP1 ÞFL (kNm) ðqF =qP2 ÞFL (kNm) 0.43 0.43 0.61 1 0.71 84.48 0 0 0 0 36.32 27.24 274.55 336.00 122.97 84.48 797.08 P qF FL qP DBH ¼ 1 P qF FL EA qP1 DBV ¼ 1 P qF FL EA qP2 ¼ 84:48 kNm EA ¼ 797:08 kNm EA ¼ 84:48 ¼ 0:00035 m 200ð10 6 Þð0:0012Þ ¼ 797:08 ¼ 0:00332 m 200ð10 6 Þð0:0012Þ DBH ¼ 0:35 mm ! Ans. DBV ¼ 3:32 mm # Ans. continued SECTION 7.8 Betti’s Law and Maxwell’s Law of Reciprocal Deflections 325 As shown in Fig. 7.24(b), a ﬁctitious horizontal force P1 ð¼ 0Þ is applied at joint B to determine the horizontal component of deﬂection, whereas the 84-kN vertical load is designated as the variable P2 to be used for computing the vertical component of deﬂection at joint B. The member axial forces, in terms of P1 and P2 , are then determined by applying the method of joints. These member forces F , along with their partial derivatives with respect to P1 and P2 , are tabulated in Table 7.13. Note that the tensile axial forces are considered as positive and the compressive forces are negative. Numerical values of P1 ¼ 0 and P2 ¼ 84 kN are then substituted in the equations for F , and the expression of Castigliano’s second theorem, as given by Eq. (7.59) is applied, as shown in the table, to determine the horizontal and vertical components of the deﬂection at joint B of the truss. 7.8 BETTI’S LAW AND MAXWELL’S LAW OF RECIPROCAL DEFLECTIONS Maxwell’s law of reciprocal deﬂections, initially developed by James C. Maxwell in 1864, plays an important role in the analysis of statically indeterminate structures to be considered in Part Three of this text. Maxwell’s law will be derived here as a special case of the more general Betti’s law, which was presented by E. Betti in 1872. Betti’s law can be stated as follows: For a linearly elastic structure, the virtual work done by a P system of forces and couples acting through the deformation caused by a Q system of forces and couples is equal to the virtual work of the Q system acting through the deformation due to the P system. To show the validity of this law, consider the beam shown in Fig. 7.25. The beam is subjected to two di¤erent systems of forces, P and Q systems, as shown in Fig. 7.25(a) and (b), respectively. Now, let us assume that we subject the beam that has the P forces already acting on it (Fig. 7.25(a)) to the deﬂections caused by the Q system of forces (Fig. 7.25(b)). The virtual external work (Wve ) done can be written as Wve ¼ P1 D Q1 þ P2 D Q2 þ þ Pn D Qn or Wve ¼ n P Pi D Qi i¼1 By applying the principle of virtual forces for deformable bodies, Wve ¼ Wvi , and using the expression for the virtual internal work done in beams (Eq. (7.29)), we obtain ðL n P MP MQ dx (7.66) Pi D Qi ¼ EI 0 i¼1 FIG. 7.25